已知[(x2+y2)-(x-y)2+2y(x-y)]÷4y=1,求4x4x
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已知[(x2+y2)-(x-y)2+2y(x-y)]÷4y=1,求
4x |
4x
[(x2+y2)-(x-y)2+2y(x-y)]÷4y
=(x2+y2-x2+2xy-y2+2xy-2y2)÷4y =(4xy-2y2)÷4y =x- 1 2y ∵x- 1 2y=1, ∴2x-y=2, ∴ 4x 4x2−y2- 1 2x+y, = 4x (2x+y)(2x−y)- 1 2x+y = 4x−2x+y (2x+y)(2x−y) = 2x+y (2x+y)(2x−y) = 1 2x−y = 1 2.
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